Degree of FreedomEnergy EquationsVirtual Work

k (N/m) c (Ns/m)

Static Equilibrium Position

kmg

static =δ

f(t)

x(t)

m

F(t) input

X(t) output

1 DOF System (Mass&Spring)

MECHANICAL SYSTEMS

k (N/m)

f(t)

m

x(t)c (Ns/m)

21 2

1 xmE &=

22 2

1 kxE =

xxcxtfW δ−δ=δ &)(

Lagrange’s Equation:

ixii

QxE

xE

dtd

=∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂ 21&

Lagrange’s equation is written for linear systems as follow

Where xi is the ith generalized co-ordinate and Qxi is the generalizedforce subjected to the ith generalized co-ordinate.

ixii

Qx

EEx

EEdtd

=∂−∂

−⎟⎟⎠

⎞⎜⎜⎝

⎛∂−∂ )()( 2121&

For Mass&spring system, application of the Lagrange’s equation givesthe “Equation of Motion”.

( ) xmxmdtd

xE

dtd

&&&&

==⎟⎠⎞

⎜⎝⎛∂∂ 1

kxx

E=

∂∂ 2

xctfQx &−= )(

xctfkxxm &&& −=+ )(

)(tfkxxcxm =++ &&&

Viscous damping, B (Nms/rad)

R

k

c

221 mRI =

m M(t)

θ

RθR Sinθ

For θ<<1

Sin(θ)=θ

Cos(θ)=1

21 2

1θ= &IE

22 )(

21

θ= RkE

)R(cRB)t(MW θδθ−δθθ−δθ=δ &&

≈R

221 2

121

θ= &mRE

222 2

1θ= kRE 1 DOF system, θ

Lagrange’s equation for θ

θ=⎟⎠⎞

⎜⎝⎛ θ=⎟

⎠⎞

⎜⎝⎛

θ∂∂ &&&&

22121

21 mRmR

dtdE

dtd θ=

θ∂∂ 22 kRE

θ−θ−=θ&& 2cRB)t(MQ

( ) )t(MkRcRBmR21 222 =θ+θ++θ &&&

)R(cRB)t(MW θδθ−δθθ−δθ=δ &&

1 DOF system, θL

kc

θF(t)

L/4 3L/8 3L/8

GA B

D

21 2

1θ= &BIE

Homogeneous beam is pinned at point B. The moment of inertia with respect to point B is calculated from theSteiner Theorem.

2mrII GB +=

r

442LLLr =−=

22

4121

⎟⎠⎞

⎜⎝⎛+=

LmmLIB

C

22

4121

⎟⎠⎞

⎜⎝⎛+=

LmmLIB 487

483

484

16121 2222

2 mLmLmLLmmLIB =+=+=

2221 48

721

21

θ=θ= && mLIE B 1621

421)(

21 222

22

θ=⎟

⎠⎞

⎜⎝⎛ θ==

LkLkxkE A

DDC xxcxtFW δ−δ=δ &)( θ=θ=8

6,8

3 LxLx DC

⎟⎠⎞

⎜⎝⎛ θδθ−⎟

⎠⎞

⎜⎝⎛ θδ=δ

86

86

83)( LLcLtFW & δθθ−δθ−=δ &

6436

83)(

2LcLtFW

δθ⎟⎟⎠

⎞⎜⎜⎝

⎛θ−−=δ &

6436

83)(

2LcLtFW

Lagrange’s equation for θ

2221 48

721

21

θ=θ= && mLIE B 1621

421)(

21 222

22

θ=⎟

⎠⎞

⎜⎝⎛ θ==

LkLkxkE A

δθ⎟⎟⎠

⎞⎜⎜⎝

⎛θ−−=δ &

6436

83)(

2LcLtFW

θ=⎟⎠⎞

⎜⎝⎛ θ=⎟

⎠⎞

⎜⎝⎛

θ∂∂ &&&&

221487

487 mLmL

dtdE

dtd

θ=θ∂

∂16

22 LkE

θ−=θ&

169)(

83 2cLtFLQ

)(161

169

487

83222 tFkLcLmL L−=θ+θ+θ &&&

ELECTRICAL SYSTEMS

L R

V1C V2

q&

L-R-C Circuit

L

q&

dtqdLVV ba&

=−

a b

(Henry)

R

q&

(Ohm)

a b

qRVV ba &=−

b

q&a C

(Farad)

qC

VV ba1

=−

According to the Kirschhoff’s Rule,

2111 VqC

andVqC

qRdtqdL ==++ &&

Where V1 denotes the input voltage and V2 denotes the output voltage.

The energy and virtual work equations for L-R-C circuit are

21 2

1 qLE &= 22 2

1 qC

E = qqRqVW δ−δ=δ &1

There is only one current in the circuit and then the system is said to be 1 DOF. The generalized co-ordinate is the charge q.

Application of the Lagrange’s equation to charge q establishes themathematical model.

( ) qLqLdtd

qE

dtd

&&&&

==⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂ 1

21 2

1 qLE &= 22 2

1 qC

E = qqRqVW δ−δ=δ &1

qCq

E 12 =∂∂ qRVQq &−= 1

11 VqC

qRqL =++ &&& 21 VqC

and =

Laplace Transformation (Kuo, 1995)

The Laplace transform is one of the mathematical tools used for thesolution of linear ordinary differential equations. In comparison with theclassical method of solving linear differential equations, the Laplacetransform method has the following two attractive features:

1.The homogeneous equation and the particular integral of the solutionare obtained in one operation.

2.The Laplace transform converts the differential equation into an algebraic equation in s. It is then possible to manipulate the algebraicequation by simple algebraic rules to obtain the solution in the s-domain. The final solution is obtained by taking the inverse Laplacetransform.

Definition of the Laplace Transform

Given the real function f(t) that satisfies the condition

∫∞ σ− ∞<0

)( dtetf t

For some finite real σ, the Laplace transform of f(t) is defined as

∫∞ −=0

)()( dtetfsF st

The variable s is referred to as the Laplace operator which is a complex variable; that is,

ω+σ= is

Example 1.

Let f(t) be a unit-step function that is defined as

⎩⎨⎧

<>=

0001)(

)(tttu

tf

The Laplace transform of f(t) is obtained as

[ ]s

es

dtetutuLsF stst 11)()()(00=−===

∞−∞ −∫

Example 2.

Consider the exponential function

tetf α−=)(

where α is a real constant

α+=−==

∞α+−

∫∞

α+−α−

sdteesF s

estt ts 1)(0

)(

0

Important Theorems of the Laplace Transform

Theorem 1: Multiplication by a Constant

Let k be a constant and F(s) be the Laplace transform of f(t). Then

[ ] )()( skFtkfL =Theorem 2: Sum and Difference

Let F1(s) and F2(s) be the Laplace transform of f1(t) and f2(t), respectively. Then

[ ] )()()()( 2121 sFsFtftfL ±=±Theorem 3: Differentiation

)0()()(lim)()(0

fssFtfssFdt

tdfLt

−=−=⎥⎦⎤

⎢⎣⎡

→

)0(...)0()0()()( )1(21 −−− −−′−−=⎥⎥⎦

⎤

⎢⎢⎣

⎡ nnnnn

ffsfssFsdt

tfdL

Theorem 4: Integration

ssFdfL

t )()(0

=⎥⎦⎤

⎢⎣⎡ ττ∫

Theorem 5: Initial -Value Theorem

)(lim)(lim0

ssFtfst ∞→→

=

Theorem 6: Final -Value Theorem

)(lim)(lim0

ssFtfst →∞→

=

)(tfkxxcxm =++ &&&

Apply the Laplace transform to the equation of motion of the mass-springsystem

[ ] [ ] )()()0()()0()0()(2 sFskXxssXcxsxsXsm =+−+−− &

[ ] ( ) )()0()0(1)(2 sFxxssXkcsms =−+−++ &

[ ] )()(2 sFsXkcsms =++

kcsmssFsXsH

++== 2

1)()()(

kcsms ++21F(s) X(s)

)(1)( 2 sFkcsms

sX++

=

(Transfer Function)

All initial conditions are assumed to be zero while evaluating the Transfer Function.

k (N/m)

f(t)

m

x(t)c (Ns/m)

m=200 kgc=300 Ns/mk=5000 N/m

f(t)=50 N (step function)

t (sec)

f(t)

50

ssF 50)( =

MATLAB Code

clc;clear;

m=200;

c=300;

k=5000;

pay=[1];

payda=[m,c,k];

step(pay*50,payda)

skcsmssX 501)( 2 ++=

0 1 2 3 4 5 6 7 80

0.002

0.004

0.006

0.008

0.01

0.012

0.014

0.016

0.018Step Response

Time (sec)

Ampl

itude

Steady-state value

x(t)

msss

sssXtxxst

ss 01.050005050

50003002001)(lim)(lim 20

==++

===→∞→